2
$\begingroup$

So I want to find a degree $7$ polynomial whose Galois group is cyclic of order $7$. I know that to do this, I need to find a polynomial which is irreducible mod every prime (not dividing the discriminant). But this is where I'm stuck, since I don't know how to prove that any particular polynomial is irreducible mod every prime. Could I have some hints?

$\endgroup$
  • $\begingroup$ Balarka, the problem is that $\phi(n)$ is even unless $n=1$ or $2$, so $\phi(n)\neq 7$, so I don't think I can find a cyclotomic field. @John Most of the irreducibility criteria I know are for over $\mathbb Q$ or for polynomials of small degree. I also know that, for a particular prime, how to construct irreducible polynomials of a certain degree, but I'm not sure how to do it for all primes simultaneously. $\endgroup$ – Nishant Jun 16 '14 at 18:09
  • $\begingroup$ Yes, I know that, but I need irreducibility over $F_p$, not just over $\mathbb Q$. $\endgroup$ – Nishant Jun 16 '14 at 18:10
  • $\begingroup$ Hmm, since $\phi(49)=42$, would the minimal polynomial of some element of an index $6$, degree $74$ subfield of $\mathbb Q(\zeta_{49})$ work? $\endgroup$ – Nishant Jun 16 '14 at 18:25
6
$\begingroup$

Let $\zeta = \exp(2\pi i/29)$ be a primitive 29th root of unity. Then the field $\mathbb Q(\zeta)/\mathbb Q$ has Galois group $$ G = (\mathbb Z/29\mathbb Z)^\times \simeq \mathbb Z/28 \mathbb Z \simeq \mathbb Z/4 \mathbb Z \times \mathbb Z/7 \mathbb Z . $$ Since 2 is a primitive root mod 29 (a generator of $(\mathbb Z/29\mathbb Z)^\times$), this Galois group is generated by the automorphism $\zeta \mapsto \zeta^2$. Furthermore, an element of order 4 can by found by the seventh power of this automorphism, namely $\zeta \mapsto \zeta^{2^7} = \zeta^{12}$. Let $H$ be the subgroup of $G$ generated by $\zeta \mapsto \zeta^{12}$, which is cyclic of order 4. Then $G/H \simeq \mathbb Z/7\mathbb Z$ is the Galois group of the fixed field $K := \mathbb Q(\zeta)^H$ over $\mathbb Q$.

Now all we have to do is find the minimal polynomial of a generator for $K/\mathbb Q$. Since $\zeta$ generates $\mathbb Q(\zeta)/\mathbb Q$, the traces $\operatorname{Tr}_H(\zeta^i)$ generate $K/\mathbb Q$. Let $$ \xi = \operatorname{Tr}_H(\zeta) = \sum_{\sigma \in H} \sigma(\zeta) = \zeta + \zeta^{12} + \zeta^{28} + \zeta^{17} . $$ By solving a large system of linear equations involving the powers of $\xi$ (or using a computer to do so), we find that $\xi$ has minimal polynomial $$ f(x) = x^7 + x^6 - 12x^5 - 7x^4 + 28x^3 + 14x^2 - 9x + 1 . $$

Remark: I chose 29 because it is the smallest prime which is 1 mod 7.

Your comment above about needing the polynomial to be reducible mod every prime except those dividing the discriminant is not true. The only primes dividing the discriminant of $f$ are 17 and 29, but $f$ splits completely mod 59 or 233, for example. This is not a coincidence. These primes are both 1 mod 29, and so they split completely in $\mathbb Q(\zeta)$, and hence also in $K$

$\endgroup$
  • $\begingroup$ (+1)In general this is the inverse galois construction that can be generalized to $\Bbb Z/n\Bbb Z$. $\endgroup$ – Balarka Sen Jun 16 '14 at 18:35
  • $\begingroup$ Awesome, thanks! But why do we know that the trace generates $K$? $\endgroup$ – Nishant Jun 16 '14 at 18:38
  • $\begingroup$ @Nishant See Bruno Joyal's answer to this question. $\endgroup$ – Dane Jun 16 '14 at 18:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.